Codeforces Round #468 (Div. 2, based on Technocup 2018 Final Round) D. Peculiar apple-tree
In Arcady‘s garden there grows a peculiar apple-tree that fruits one time per year. Its peculiarity can be explained in following way: there are n inflorescences, numbered from 1 to n
Once tree starts fruiting, there appears exactly one apple in each inflorescence. The same moment as apples appear, they start to roll down along branches to the very base of tree. Each second all apples, except ones in first inflorescence simultaneously roll down one branch closer to tree base, e.g. apple in a
Help Arcady with counting number of apples he will be able to collect from first inflorescence during one harvest.
InputFirst line of input contains single integer number n (2?≤?n?≤?100?000) — number of inflorescences.
Second line of input contains sequence of n?-?1 integer numbers p2,?p3,?...,?pn (1?≤?pi?<?i), where pi is number of inflorescence into which the apple from i-th inflorescence rolls down.
OutputSingle line of output should contain one integer number: amount of apples that Arcady will be able to collect from first inflorescence during one harvest.
Examples Input Copy3Output
1 1
1Input Copy
5Output
1 2 2 2
3Input Copy
18Output
1 1 1 4 4 3 2 2 2 10 8 9 9 9 10 10 4
4Note
In first example Arcady will be able to collect only one apple, initially situated in 1st inflorescence. In next second apples from 2nd and 3rd inflorescences will roll down and annihilate, and Arcady won‘t be able to collect them.
In the second example Arcady will be able to collect 3 apples. First one is one initially situated in first inflorescence. In a second apple from 2nd inflorescence will roll down to 1st (Arcady will collect it) and apples from 3rd, 4th, 5th inflorescences will roll down to 2nd. Two of them will annihilate and one not annihilated will roll down from 2-nd inflorescence to 1st one in the next second and Arcady will collect it.
分析:
因為只有同一時間在同一地點的蘋果相碰會消失,那麽相碰一定只出現在同一層的蘋果之間,且在該層蘋果大於1的時候,一定會出現相碰。
所以我們可以將蘋果分層進行討論
如果該層的蘋果數為奇數,那麽無論經過怎樣的過程,到最後始終會剩一個。
如果為偶數,那麽到最後一定一個也不剩。
將每層得到的結果累加即可
#include <cstdio> #include <iostream> #include <algorithm> #include <vector> using namespace std; typedef long long LL; vector<int>V[100010]; int num[100010]; int maxx; void dfs(int id,int k) { num[k]++; maxx=max(k,maxx); for(int i=0;i<V[id].size();i++) dfs(V[id][i],k+1); } int main() { int n,x,ans; cin>>n; maxx=0; ans=0; for(int i=2;i<=n;i++) { cin>>x; V[x].push_back(i); } dfs(1,1); for(int i=1;i<=maxx;i++) { if(num[i]%2==1) ans++; } cout<<ans<<endl; return 0; }
Codeforces Round #468 (Div. 2, based on Technocup 2018 Final Round) D. Peculiar apple-tree